Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>

using namespace std;

using lint = long long;
using ldouble = long double;

const int INF = 1 << 30;

vector<int> bfs(const vector<vector<int>>& graph, int s, int t) {
    vector<int> dist(graph.size(), INF);
    dist[s] = 0;

    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int v = que.front();
        que.pop();

        for (int sv : graph[v]) {
            if (dist[sv] <= dist[v] + 1) continue;
            dist[sv] = dist[v] + 1;
            que.push(sv);
        }
    }
    return dist;
}

int main() {
    int N, M, s, t;
    cin >> N >> M >> s >> t;
    --s, --t;

    vector<vector<int>> graph(N);
    for (int i = 0; i < M; ++i) {
        int u, v;
        cin >> u >> v;
        --u, --v;

        graph[u].push_back(v);
        graph[v].push_back(u);
    }

    auto dist = bfs(graph, s, t);
    int D = dist[t];

    vector<lint> num(N, 0);
    for (int v = 0; v < N; ++v) {
        if (dist[v] < INF) ++num[dist[v]];
    }

    dist = bfs(graph, t, s);
    vector<lint> rnum(N, 0);
    for (int v = 0; v < N; ++v) {
        if (dist[v] < INF) ++rnum[dist[v]];
    }

    lint ans = 0;
    for (int d = 0; d <= D - 2; ++d) {
        ans += num[d] * rnum[D - d - 2];
    }

    cout << ans << endl;
    return 0;
}